高数三道大题求解析17.化简sin(2π-α)cos(π/2+α)cos(π/2-α)/[sin(3π-α)sin(-π-α)sin(π/2+α)]=多少?18.若cos(α+π/3)=4/5,α∈(0,π/2),则cosα=多少?19.已知函数f(x)=1+√2cos(2x-π/4)/[sin(x+π/2)](1)求f(x)的定义
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![高数三道大题求解析17.化简sin(2π-α)cos(π/2+α)cos(π/2-α)/[sin(3π-α)sin(-π-α)sin(π/2+α)]=多少?18.若cos(α+π/3)=4/5,α∈(0,π/2),则cosα=多少?19.已知函数f(x)=1+√2cos(2x-π/4)/[sin(x+π/2)](1)求f(x)的定义](/uploads/image/z/1822626-18-6.jpg?t=%E9%AB%98%E6%95%B0%E4%B8%89%E9%81%93%E5%A4%A7%E9%A2%98%E6%B1%82%E8%A7%A3%E6%9E%9017.%E5%8C%96%E7%AE%80sin%282%CF%80-%CE%B1%29cos%28%CF%80%2F2%2B%CE%B1%29cos%28%CF%80%2F2-%CE%B1%29%2F%5Bsin%283%CF%80-%CE%B1%29sin%28-%CF%80-%CE%B1%29sin%28%CF%80%2F2%2B%CE%B1%29%5D%3D%E5%A4%9A%E5%B0%91%3F18.%E8%8B%A5cos%28%CE%B1%2B%CF%80%2F3%29%3D4%2F5%2C%CE%B1%E2%88%88%EF%BC%880%2C%CF%80%2F2%EF%BC%89%2C%E5%88%99cos%CE%B1%3D%E5%A4%9A%E5%B0%91%3F19.%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D1%2B%E2%88%9A2cos%282x-%CF%80%2F4%29%2F%5Bsin%28x%2B%CF%80%2F2%29%5D%281%29%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%AE%9A%E4%B9%89)
高数三道大题求解析17.化简sin(2π-α)cos(π/2+α)cos(π/2-α)/[sin(3π-α)sin(-π-α)sin(π/2+α)]=多少?18.若cos(α+π/3)=4/5,α∈(0,π/2),则cosα=多少?19.已知函数f(x)=1+√2cos(2x-π/4)/[sin(x+π/2)](1)求f(x)的定义
高数三道大题求解析
17.化简sin(2π-α)cos(π/2+α)cos(π/2-α)/[sin(3π-α)sin(-π-α)sin(π/2+α)]=多少?
18.若cos(α+π/3)=4/5,α∈(0,π/2),则cosα=多少?
19.已知函数f(x)=1+√2cos(2x-π/4)/[sin(x+π/2)]
(1)求f(x)的定义域;(2)若角α是第四象限角,且cosα=3/5,求f(α).
高数三道大题求解析17.化简sin(2π-α)cos(π/2+α)cos(π/2-α)/[sin(3π-α)sin(-π-α)sin(π/2+α)]=多少?18.若cos(α+π/3)=4/5,α∈(0,π/2),则cosα=多少?19.已知函数f(x)=1+√2cos(2x-π/4)/[sin(x+π/2)](1)求f(x)的定义
17.化简sin(2π-α)cos(π/2+α)cos(π/2-α)/[sin(3π-α)sin(-π-α)sin(π/2+α)]=多少?
=(-sinα)(-sinα)sinα/[sinα(-sinα)cosα)=-tanα
18.若cos(α+π/3)=4/5,α∈(0,π/2),则cosα=多少?
∵α∈(0,π/2),∴α+π/3∈(π/3,5π/6)
∵cos(α+π/3)=4/5 ∴sin(α+π/3)=3/5
∴cosα=cos[(α+π/3)-π/3]
=cos(α+π/3)cosπ/3+sin(α+π/3)sinπ/3
=4/5*1/2+3/5*√3/2=(4+3√3)/10
19.已知函数f(x)=1+√2cos(2x-π/4)/[sin(x+π/2)]
(1)求f(x)的定义域;(2)若角α是第四象限角,且cosα=3/5,求f(α).
(1) sin(x+π/2)=cosx≠0,定义域{x|x≠kπ+π/2,k∈Z}
(2) f(x)=(1+cos2x+sin2x)/cosx
=(2cos²x+2sinxcosx)/cosx
=2(sinx+cosx)
∵角α是第四象限角,且cosα=3/5,
∴sinα=-4/5
∴f(α)= 2(sinα+cosα)=-2/5